Problem: Simplify the following expression: $ n = \dfrac{1}{4} - \dfrac{-r - 3}{r + 2} $
Answer: In order to subtract expressions, they must have a common denominator. Multiply the first expression by $\dfrac{r + 2}{r + 2}$ $ \dfrac{1}{4} \times \dfrac{r + 2}{r + 2} = \dfrac{r + 2}{4r + 8} $ Multiply the second expression by $\dfrac{4}{4}$ $ \dfrac{-r - 3}{r + 2} \times \dfrac{4}{4} = \dfrac{-4r - 12}{4r + 8} $ Therefore $ n = \dfrac{r + 2}{4r + 8} - \dfrac{-4r - 12}{4r + 8} $ Now the expressions have the same denominator we can simply subtract the numerators: $n = \dfrac{r + 2 - (-4r - 12) }{4r + 8} $ Distribute the negative sign: $n = \dfrac{r + 2 + 4r + 12}{4r + 8}$ $n = \dfrac{5r + 14}{4r + 8}$